高中数学题4.3

1.函数y=sin2x是( ) A.周期为 的奇函数 B.周期为 的偶函数 C.周期为2 的奇函数 D.周期为2 的偶函数 【答案】 A 【解析】 EMBED Equation.DSMT4,f(-x)=sin(-2x)=-sin2x=-f(x), ∴函数是周期为 的奇函数. 2.函数 的定义域为( ) A. B.[2k EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z) C.(2k EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z) D.[k EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z) 【答案】 B 【解析】 若函数 有意义, 则2sin 即sin . 由函数y=sinx的图象(图略),得2k EMBED Equation.DSMT42k + Z, 所以函数 的定义域为[2k + EMBED Equation.DSMT4 EMBED Equation.DSMT4Z). 3.函数f(x)=cos 的单调递增区间为( ) A.[- ,0] B.[(2k-1) ,2k EMBED Equation.DSMT4Z) C.[2k EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z) D.[k EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z) 【答案】 D 【解析】 由(2k-1) EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z),解得k EMBED Equation.DSMT4 EMBED Equation.DSMT4k EMBED Equation.DSMT4Z). ∴函数f(x)的单调递增区间为[k EMBED Equation.DSMT4k + EMBED Equation.DSMT4Z). 4.（2012山东日照测试）已知函数f(x)=sin 的最小正周期为 ,则函数f(x)的图象的一条对称轴方程是( ) A. B. C. D. 【答案】 C 【解析】 由T= EMBED Equation.DSMT4得 所以f(x)=sin 则函数f(x)的对称轴为 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z,解得x= EMBED Equation.DSMT4Z,所以 为函数f(x)的一条对称轴. 5.函数y=sin sinx+1的值域是. 【答案】 【解析】 y=sin sinx+1=(sin . 当sin 时,函数取得最小值 ; 当sinx=1时,函数取得最大值3. 1.下列函数中,周期为 的是( ) A.y=sin B.y=sin2x C.y=|sinx| D.y=sin4x 【答案】 D 2.函数f(x)=tan 的图象的相邻的两支截直线 所得线段长为 则 的值是( ) A.0 B.1 C.-1 D. 【答案】 A 【解析】 由于相邻的两支截直线 所得的线段长为 所以该函数的周期 因此 函数解析式为f(x)=tan4x,所以 tan tan =0. 3.函数y=2sin EMBED Equation.DSMT4])为增函数的区间是 … ( ) A. B. C. D. EMBED Equation.DSMT4] 【答案】 C 【解析】 ∵y=2sin sin ∴y=2sin 的递增区间实际上是 u=2sin 的递减区间, 即2k EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z), 解得k EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z). 令k=0,得 . 又∵ EMBED Equation.DSMT4],∴ 即函数y=2sin EMBED Equation.DSMT4])的递增区间为 . 4.函数y=|sinx|-2sinx的值域是( ) A.[-3,-1] B.[-1,3] C.[0,3] D.[-3,0] 【答案】 B 【解析】 当 sin 时,y=sinx-2sinx=-sinx,此时 [-1,0];当 sinx<0时,y=-sinx-2sinx=-3sinx,这时 求其并集得 . 5.若函数y=2cos 在区间 上递减,且有最小值1,则 的值可以是( ) A.2 B. C.3 D. 【答案】 B 【解析】 由y=2cos 在 上是递减的,且有最小值为1,则有 即 cos cos .检验各数据,得出B项符合. 6.对于函数f(x)=|sin2x|有下列命 题 快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题 : ①函数f(x)的最小正周期是 ; ②函数f(x)是偶函数; ③函数f(x)的图象关于直线 对称; ④函数f(x)在 上为减函数. 其中正确命题的序号是( ) A.②③ B.②④ C.①③ D.①② 【答案】 D 【解析】 函数f(x)=|sin2x|的最小正周期是 且是偶函数,则①②正确,故选D. 7.(2012福建福州检测)下列说法正确的是( ) A.函数y=sin 在区间 内单调递增 B.函数y=cos sin 的最小正周期为2 C.函数y=cos 的图象是关于点 成中心对称的图形 D.函数y=tan 的图象是关于直线 成轴对称的图形 【答案】 C 【解析】 令 则 此时y=sin 不单调,故A选项为假命题;y=cos sin cos sin cos2x,最小正周期为 ,故B选项也为假命题;正切函数的图象不是轴对称图形,故排除D;当 时,cos 所以 是对称中心,故选C. 8.函数 的定义域是 . 【答案】 (k EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z) 【解析】 由1-tan 得tan ∴k EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z). 9.f(x)是以5为周期的奇函数,f(-3)=4且cos 则f(4cos . 【答案】 -4 【解析】 ∵4cos cos 又T=5,且f(x)为奇函数, ∴f(4cos 2+5) =f(3)=-f(-3)=-4. 10.设函数y=sin 若对任意 R,存在 使 恒成立,则| |的最小值是 . 【答案】 2 【解析】 由 恒成立,可得 为最小值 为最大值,| |的最小值为半个周期. 11.已知函数f(x)=log sin . (1)求函数的定义域; (2)求满足f(x)=0的x的取值范围. 【解】 (1)令 sin 有sin 故2k < EMBED Equation.DSMT4+ EMBED Equation.DSMT4Z,∴k EMBED Equation.DSMT4 EMBED Equation.DSMT4 Z. 故函数f(x)的定义域为(k EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z. (2)∵f(x)=0,∴log sin . ∴ sin . ∴sin . ∴ EMBED Equation.DSMT4 EMBED Equation.DSMT4或2k EMBED Equation.DSMT4Z. 故x=k EMBED Equation.DSMT4或x=k EMBED Equation.DSMT4Z. 故x的取值范围是{x|x=k EMBED Equation.DSMT4或x=k EMBED Equation.DSMT4 Z}. 12.已知函数f(x)=cos sin sin . (1)求函数f(x)的最小正周期; (2)求函数f(x)在区间 上的值域. 【解】 (1)f(x)=cos sin sin cos sin2x+(sinx-cosx)(sinx+cosx) cos sin2x-cos2x =sin . ∴周期为 EMBED Equation.DSMT4. (2)∵ ∴ . ∵f(x)=sin 在区间 上单调递增,在区间 上单调递减, ∴当 时,f(x)取得最大值1. 又∵ ∴当 时,f(x)取得最小值 . ∴函数f(x)在 上的值域为 . 13.是否存在实数a,使得函数y=sin cos 在闭区间 上的最大值是1?若存在,求出对应的a值;若不存在,请说明理由. 【解】 y=1-cos cos =-(cos . ∵ ∴ cos . ①若 即a>2,则当cosx=1时, 舍去); ②若 即 则当cos 时 ∴ 或a=-4<0(舍去); ③若 即a<0,则当cosx=0时, 舍去). 综上,知存在 符合题意. 14.已知函数f(x)=sin2x+acos R,a为常数),且 是函数y=f(x)的零点. (1)求a的值,并求函数f(x)的最小正周期; (2)若 求函数f(x)的值域,并写出f(x)取得最大值时x的值. 【解】 (1)由于 是函数y=f(x)的零点, 即 是方程f(x)=0的解, 从而 sin cos 则 解得a=-2. 所以f(x)=sin2x-2cos sin2x-cos2x-1, 则 sin 所以函数f(x)的最小正周期为 . (2)由 得 则sin 则 sin sin ∴函数f(x)的值域为 . 当 EMBED Equation.DSMT4 EMBED Equation.DSMT4Z), 即x=k EMBED Equation.DSMT4Z)时,f(x)有最大值, 又 故k=0时 f(x)有最大值 . _1385898846.unknown _1385899494.unknown _1385899920.unknown _1385900328.unknown _1385900516.unknown _1385900575.unknown _1385900626.unknown _1385900727.unknown _1385900729.unknown _1385900730.unknown _1385900728.unknown _1385900725.unknown _1385900726.unknown _1385900723.unknown _1385900724.unknown _1385900632.unknown _1385900597.unknown _1385900608.unknown _1385900615.unknown _1385900621.unknown _1385900602.unknown _1385900585.unknown _1385900592.unknown _1385900580.unknown _1385900550.unknown _1385900563.unknown _1385900570.unknown _1385900556.unknown _1385900537.unknown _1385900544.unknown _1385900524.unknown _1385900532.unknown _1385900372.unknown _1385900488.unknown _1385900502.unknown _1385900508.unknown _1385900494.unknown _1385900472.unknown _1385900480.unknown _1385900376.unknown _1385900351.unknown _1385900362.unknown _1385900366.unknown 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